BZOJ4643 卡常大水题 【Tarjan】

题目分析：

给所有边按A排序，依次加入再按B递增排序，势能分析可以发现是O(n^4)的

代码：

#include<bits/stdc++.h>
using namespace std;
 
const int maxn = ;
const int maxm = ;
 
int n,cl,dfn[maxn],low[maxn];
struct edge{int u,v,w1,w2;}edges[maxm];
vector <int> g[maxn];
int cmp(edge alpha,edge beta){return alpha.w1 < beta.w1;}
 
set<pair<int,pair<int,int> > > st;
 
void Tarjan(int now){
    low[now] = dfn[now] = ++cl;
    for(int i=;i<g[now].size();i++){
        int t = g[now][i];
        if(dfn[t]) low[now] = min(low[now],dfn[t]);
        else{Tarjan(t); low[now] = min(low[now],low[t]);}
    }
}
 
int chk(){// to check if it is Strong Connect Graph
    cl = ;
    for(int i=;i<=n;i++) low[i] = dfn[i] = ;
    Tarjan();
    for(int i=;i<=n;i++) if(low[i] == dfn[i]) return ;
    return ;
}
 
void read(){
    scanf("%d",&n);
    for(int i=;i<=n;i++) for(int j=;j<=n;j++){
            int x; scanf("%d",&x);
            edges[n*(i-)+j]=(edge){i,j,x,};
        }
    for(int i=;i<=n;i++) for(int j=;j<=n;j++) {
            int x; scanf("%d",&x);
            edges[n*(i-)+j].w2 = x;
        }
    sort(edges+,edges+n*n+,cmp);
}
 
void work(){
    int ans = 2e9;
    int flag = ,lstans = ;
    for(int i=;i<=n*n;i++){
        if(edges[i].u == edges[i].v) continue;
        int bb = edges[i].w1;
        if(flag&&lstans < edges[i].w2) continue;
        st.insert(make_pair(edges[i].w2,make_pair(edges[i].u,edges[i].v)));
        g[edges[i].u].push_back(edges[i].v);
        if(!flag){
            if(chk()){
                set<pair<int,pair<int,int> > >::iterator it = st.end();
                it--;
                lstans = (*it).first;
                flag = ;
                ans = min(ans,lstans+bb);
            }
            continue;
        }
        while(true){
            set<pair<int,pair<int,int> > >::iterator it = st.end();
            it--;int u = (*it).second.first,v = (*it).second.second;
            for(int i=;i<g[u].size();i++){
                if(g[u][i] == v){
                    swap(g[u][i],g[u][g[u].size()-]);
                    g[u].pop_back();
                    break;
                }
            }
            if(chk()){
                st.erase(it);
            }else {g[u].push_back(v);break;}
        }
        set<pair<int,pair<int,int> > >::iterator it = st.end();
        it--;lstans = (*it).first;
        ans = min(ans,(*it).first+bb);
    }
    printf("%d\n",ans);
}
 
int main(){
    read();
    work();
    return ;
}